Question

drive the a formula for root mean square value of alternating current.
(or)
prove that Irms=Io/√2

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Answer 1

$use \: \: \: \boxed{I = I_{o}sin \: \omega \: t }$

Explanation:

$we \: \: know \: \: that \: \: \boxed{I = I_{o}sin \: \omega \: t } \\ Now, \: \: \: I_{rms}^{2} = \frac{ \int_{0 }^{T} {I}^{2}dt }{ \int_{0 }^{T}dt} \\ = \frac{ \int_{0 }^{T} {(I_{o}sin \: \omega \: t})^{2}dt }{T} \\ = \frac{ \int_{0 }^{T} {I_{o}^{2} sin^{2} \: \omega \: t}dt }{T} \\ = \frac{I_{o}^{2}}{2T} { \int_{0 }^{T}{ (1 - cos2 \omega \: t)}dt } \\ = \frac{I_{o}^{2}}{2T} [ t - \frac{sin2 \omega \: t}{2 \omega} ]_{T}^{0} \\ = \frac{I_{o}^{2}}{2T}\times [(0 - 0) - (0 - T)] \\ = \frac{I_{o}^{2}}{2T} \times T \\ =\frac{I_{o}^{2}}{2} \\ so \: \: \: \: I_{rms} = \sqrt{\frac{I_{o}^{2}}{2} } = \frac{I_{o}}{ \sqrt{2} }$